I've been following a series on understanding about how imaginary numbers came to be, and in the series, it mentions that imaginary numbers mostly follows the algebra rules for real numbers, such as adding or multiplying by real numbers. However, it specifically mentions this "inconsistency(?)" about multiplying square roots of imaginary numbers do not follow the rule for multiplying square roots of real numbers, namely $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$. For example, why can't I do this: $\sqrt{-2} \times \sqrt{-3} = \sqrt{(-2)(-3)} = \sqrt{6}$, which I know is the wrong answer. In the past, I've just memorized to factor out the imaginary parts first, and that $i^2 = -1.$ However, can anyone show me an explanation for this, or tell me where I could learn more about it?

  • $\begingroup$ Because $x \mapsto \sqrt{x}$ has two branches. $\endgroup$ – DHMO Apr 16 '17 at 13:44
  • $\begingroup$ Do you mind elaborating? Also, I do not have a very good understanding of math, since I'm only a sophomore from an engineering major... $\endgroup$ – Tony Tarng Apr 16 '17 at 13:47
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    $\begingroup$ Actually it does work. Simply, $\sqrt{a}$ is not well defined in complex numbers. So the precise theorem is : "if $c$ is a square root of $a$, $d$ a square root of $b$, then $cd$ is a square root of $ab$" $\endgroup$ – Max Apr 16 '17 at 13:48

$\sqrt\cdot$ is well defined on $\Bbb R_+$ because $\Bbb R$ has a total order, which makes us able to make a sensible choice between the square roots of $b$ (ie $\sqrt b$ and $-\sqrt b$), that is, take the positive one. However in $\Bbb C$, you no longer can compare numbers ($3i \le 5$ makes no sense, for instance).

Therefore in your example $i\sqrt 2 \times i\sqrt 3 = -\sqrt 6$ is in fact a square root of $6$, but it is not the square root of $6$ as we like to call it in $\Bbb R$. In $\Bbb R$ since the product of two positive numbers is positive, we do not have this issue.


It all comes from how we define the square root of something. In complex analysis, we have that every number can be written as $re^{i\theta} = re^{i(\theta +2\pi)}$. This comes from Euler's identity and writing the number in polar coordinates.

As a result, since $e^{i\theta}$ is periodic with period $2\pi$, when we define the logarithm of a number we need to either make it discontinuous or define it only on the plane minus a ray from the origin. We choose the second one since it makes our lives much easier.

Now, to define square roots in the complex plane, we let $\sqrt{z}=e^{\frac{\log(z)}{2}}$. Now again since $e^{i\theta}$ has period $2\pi$, the identity $\sqrt{zw}=\sqrt{z}\sqrt{w}$ is true modulo $2\pi$ of the angle. Also the ray you remove from the origin needs to be the same for all the terms involved to have this identity, which for positive real numbers is the negative real axis, and for negative real numbers is the positive real axis usually. So another problem is you are switching what is called the branch of your logarithm.

  • $\begingroup$ you mean $re^{i(\theta+2n\pi)}$ $\endgroup$ – DHMO Apr 16 '17 at 13:57
  • $\begingroup$ Or more simply, if $a$ is a square root of $b$, then $-a$ is also a square root of $b$. $\endgroup$ – DHMO Apr 16 '17 at 13:58
  • $\begingroup$ @DHMO sure, but that follows from the case when n=1. $\endgroup$ – user416426 Apr 16 '17 at 13:58

First thing to note -

${\sqrt i}^2 \ne -1$

It is $i = \sqrt {-1}$ and $i^2 = -1$

Now you can see -

$\sqrt {-2} × \sqrt {-3} = \sqrt 2 × \sqrt 3$

  • $\begingroup$ Ooops, I meant $i^2 = -1$. $\endgroup$ – Tony Tarng Apr 16 '17 at 14:20

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